A capacitor is made from two hollow, coaxial, iron cylinders, one inside the other. The inner cylinder is negatively charged and the outer is positively charged; the magnitude of the charge on each is 11.5 pC. The inner cylinder has a radius of 0.550 mm, the outer one has a radius of 4.00 mm, and the length of each cylinder is 15.0 cm.

(1) What is the capacitance? Use 8.854×10−12 F/m for the permittivity of free space.
(2) What applied potential difference is necessary to produce these charges on the cylinders?

Given the permittivity constant to be [tex]8.854\times 10^-^1^2 F/m[/tex],The capacitance of a [tex]cylindrical \ capacitor[/tex] of length, [tex]L[/tex] is given by the equation:

[tex]C=\frac{2\pi \epsilon _o L}{ln(b/a)}[/tex] where [tex]b[/tex] is the radius of the outer cylinder and [tex]a[/tex] the radius of the inner cylinder.

The values are given as:[tex]a=0.550mm(5.5\times 10^-^4m), \ b=4.00mm(4.0\times10^-^3m), \ L=15.0cm(0.150m)[/tex]

Substitute in our capacitance equation:

[tex]C=\frac{2\times\pi \times 8.854\times 10^-^1^2 \times 0.15}{In(4.00/0.550)}\\=4.2057\times 10^-^1^2 F[/tex]

Hence the capacitance is [tex]4.2057\times 10^-^1^2 F[/tex]

b. The charge on the capacitance is related to the potential difference across it. The potential difference is expressed using the equation:

[tex]Q=CV[/tex],[tex]Q=11.5pC[/tex]

From a above, we already have our capacitance value,[tex]C=4.2057\times 10^-^1^2 F[/tex]

We substitute [tex]C[/tex] in the pd equation:

[tex]v=>(11.5)/(4.2057)\\=2.7344V[/tex]

Hence, the applied potential difference is 2.7344V